Science:Math Exam Resources/Courses/MATH152/April 2012/Question 01 (c)/Solution 1

From 1(b) we saw that the system of equations the line must satisfy is given by

$\begin{array}{rl}2x_1+x_2-x_3& =1\\ x_1+x_2& =1.\end{array}$

In order to find the parametric form we solve this system. First we put it in augmented form

$\left[\begin{array}{cccc}2& 1& -1& 1\\ 1& 1& 0& 1\end{array}\right]$

and then perform row operations to reduce the matrix. First we will swap row 1 and row 2 so that the first pivot (row 1, column 1) has value 1.

$\left[\begin{array}{cccc}1& 1& 0& 1\\ 2& 1& -1& 1\end{array}\right]$

and then we will subtract 2 multiples of row 1 from row 2 to place a zero below the pivot,

$\left[\begin{array}{cccc}1& 1& 0& 1\\ 0& -1& -1& -1\end{array}\right].$

We will then multiply the second row by -1 to put a 1 in the pivot position there (row 2, column 2),

$\left[\begin{array}{cccc}1& 1& 0& 1\\ 0& 1& 1& 1\end{array}\right]$

and finally we will subtract 1 multiple of row 2 from row 1 to put a 0 above the pivot,

$\left[\begin{array}{cccc}1& 0& -1& 0\\ 0& 1& 1& 1\end{array}\right].$

Notice that because we have more columns than rows, then we have a free variables (the matrix is rank deficient). Let the free variable be $x_3$, i.e. let $x_3=t$. Then from our reduced matrix problem we have that $x_1=t$ and that $x_2+t=1$ or $x_2=1-t$. Therefore we can write that

$\left[\begin{array}{c}{x}_1\\ x_2\\ x_3\end{array}\right]=\left[\begin{array}{c}0\\ 1\\ 0\end{array}\right]+t\left[\begin{array}{c}1\\ -1\\ 1\end{array}\right].$

This parametrically describes the line.